Question

#calculus #limit
$\lim _ { x \rightarrow 0 } \frac { 3 ^ { 2 x } + 3 ^ { x + 1 } - 4 } { 3 ^ { 2 x } - 1 }$
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Answer 1

We need to evaluate the given limits:

$\longrightarrow\lim \limits_ { x \rightarrow 0 } \dfrac { 3 ^ { 2 x } + 3 ^ { x + 1 } - 4 } { 3 ^ { 2 x } - 1 }$

By directly substituting the limits, we get:

$\dfrac { 3 ^ { 2 (0) } + 3 ^ { 0 + 1 } - 4 } { 3 ^ { 2 (0) } - 1} = \dfrac{1 - 1}{1 - 1} = \boxed{ \dfrac{0}{0}}$

This is an indeterminate quantity therefore we have to use another method to solve the problem.

Solution:

$\implies\lim \limits_ { x \rightarrow 0 } \dfrac { 3 ^ { 2 x } + 3 ^ { x + 1 } - 4 } { 3 ^ { 2 x } - 1}$

$\implies\lim \limits_ { x \rightarrow 0 } \dfrac { 3 ^ { 2 x } + 3 ^ { x + 1 } - 1 - 3 } { 3 ^ { 2 x } - 1}$

${ \implies\lim \limits_ { x \rightarrow 0 } \dfrac {( 3 ^ { 2 x } - 1)+ (3 ^ { x + 1 } - 3 )} { 3 ^ { 2 x } - 1}}$

${ \implies\lim \limits_ { x \rightarrow 0 } \cancel\dfrac {( 3 ^ { 2 x } - 1)}{ {3}^{2x} - 1}+ \dfrac{ (3 ^ { x + 1 } - 3 )} { 3 ^ { 2 x } - 1}}$

${ \implies1 + \lim \limits_ { x \rightarrow 0 } \dfrac{ 3(3 ^ { x} - 1)} { (3 ^ {x })^{2} - (1)^{2} }}$

${ \implies1 + \lim \limits_ { x \rightarrow 0 } \dfrac{ 3 \cancel{(3 ^ { x} - 1)}} { (3 ^ {x } + 1) \cancel{( {3}^{x} - 1) }}}$

${ \implies1 + \lim \limits_ { x \rightarrow 0 } \dfrac{ 3} { (3 ^ {x } + 1)}}$

${ \implies1 + \dfrac{ 3} { (3 ^ {0} + 1)}}$

${ \implies1 + \dfrac{ 3} { (1+ 1)}}$

${ \implies1 + \dfrac{ 3}2}$

${ \implies \dfrac{ 5}2}$

Therefore the required answer is:

$\boxed{ \lim \limits_ { x \rightarrow 0 } \dfrac { 3 ^ { 2 x } + 3 ^ { x + 1 } - 4 } { 3 ^ { 2 x } - 1 } = \frac{5}{2} }$